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Ch. 7 Solutions

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Table of Contents

Basic

Medium

Advanced

Basic

B-1. Substituting a Plane Wave into the Free-Particle Schrödinger Equation

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Solution Strategy: Substitute the plane wave \(\Psi(x,t) = Ae^{i(kx - \omega t)}\) into the free-particle Schrödinger equation and find the relationship between \(\omega\) and \(k\).

Detailed Calculation:

Calculate the left-hand side:

\[i\hbar\frac{\partial\Psi}{\partial t} = i\hbar \cdot (-i\omega) \Psi = \hbar\omega\,\Psi\]

Calculate the right-hand side:

\[-\frac{\hbar^2}{2m}\frac{\partial^2\Psi}{\partial x^2} = -\frac{\hbar^2}{2m}\cdot(ik)^2\Psi = -\frac{\hbar^2}{2m}\cdot(-k^2)\Psi = \frac{\hbar^2 k^2}{2m}\Psi\]

Dividing both sides by \(\Psi \neq 0\):

\[\hbar\omega = \frac{\hbar^2 k^2}{2m}\]

Final Answer:

\[\boxed{\omega = \frac{\hbar k^2}{2m}}\]

Verification: Substituting \(p = \hbar k\) and \(E = \hbar\omega\) gives \(E = p^2/(2m)\), which agrees with the classical relationship between energy and momentum for a free particle. ✓

B-2. Apply the momentum operator to each of the following wave functions and find the result. If it is an eigenfunction, state the eigenvalue.

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Solution strategy: Apply \(\hat{p} = -i\hbar\frac{\partial}{\partial x}\) to each wave function and check whether the result takes the form \(\hat{p}\psi = (\text{constant})\cdot\psi\).

(a) \(\psi(x) = e^{5ix/\hbar}\)

\[\hat{p}\psi = -i\hbar\frac{\partial}{\partial x}e^{5ix/\hbar} = -i\hbar \cdot \frac{5i}{\hbar}\,e^{5ix/\hbar} = 5\,e^{5ix/\hbar} = 5\,\psi\]

This is an eigenfunction, with eigenvalue \(p = 5\) (in appropriate units).

(b) \(\psi(x) = \cos(kx)\)

\[\hat{p}\psi = -i\hbar\frac{\partial}{\partial x}\cos(kx) = -i\hbar\cdot(-k\sin(kx)) = i\hbar k\sin(kx)\]

The result is \(\sin(kx)\), which is not a constant multiple of the original \(\cos(kx)\).

This is not an eigenfunction. The result is \(\hat{p}\cos(kx) = i\hbar k\sin(kx)\).

(c) \(\psi(x) = (x^2 + 1)e^{ipx/\hbar}\)

Using the product rule:

\[\hat{p}\psi = -i\hbar\frac{\partial}{\partial x}\left[(x^2+1)e^{ipx/\hbar}\right]\]
\[= -i\hbar\left[2x\,e^{ipx/\hbar} + (x^2+1)\cdot\frac{ip}{\hbar}\,e^{ipx/\hbar}\right]\]
\[= -2i\hbar x\,e^{ipx/\hbar} + p(x^2+1)e^{ipx/\hbar}\]
\[= \left[p(x^2+1) - 2i\hbar x\right]e^{ipx/\hbar}\]

The result is not a constant multiple of the original \(\psi = (x^2+1)e^{ipx/\hbar}\) (an \(x\)-dependent coefficient remains).

This is not an eigenfunction.

Verification: (a) has the form \(e^{ipx/\hbar}\) (with \(p=5\)), so it should be a momentum eigenfunction. ✓ (b) is a superposition of \(e^{ikx}\) and \(e^{-ikx}\), so it does not have a definite momentum. ✓

B-3. Normalize the wave function (where is a constant). That is

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Solution strategy: Integrate \(|\Psi|^2\) over all space and apply the Gaussian integral formula to find \(A\).

Detailed calculation:

\[\int_{-\infty}^{+\infty}|\Psi(x)|^2\,dx = A^2\int_{-\infty}^{+\infty}e^{-x^2/a^2}\,dx = 1\]

Setting \(\alpha = 1/a^2\) and using the Gaussian integral formula:

\[\int_{-\infty}^{+\infty}e^{-x^2/a^2}\,dx = \sqrt{\frac{\pi}{1/a^2}} = \sqrt{\pi a^2} = a\sqrt{\pi}\]

Therefore:

\[A^2 \cdot a\sqrt{\pi} = 1\]
\[A^2 = \frac{1}{a\sqrt{\pi}}\]

Final answer:

\[\boxed{A = \frac{1}{(a\sqrt{\pi})^{1/2}} = \left(\frac{1}{a\sqrt{\pi}}\right)^{1/2} = \frac{1}{\pi^{1/4}\sqrt{a}}}\]

Verification: Dimensional analysis: \(A^2\) should have dimensions of \([1/\text{length}]\) (since \(|\Psi|^2\) is a probability density with dimensions \([1/\text{length}]\)). \(A^2 = 1/(a\sqrt{\pi})\) has dimensions of \([1/\text{length}]\). ✓

B-4. Properties of the Dirac Delta Function

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Solution strategy: Directly apply the property \(\int f(x)\delta(x - x_0)dx = f(x_0)\).

(a)

\[\int_{-\infty}^{+\infty}(3x^2 + 2)\,\delta(x-1)\,dx = 3(1)^2 + 2 = \boxed{5}\]

(b)

\[\int_{-\infty}^{+\infty}e^{ikx}\,\delta(x)\,dx = e^{ik\cdot 0} = \boxed{1}\]

(c)

\[\int_{-\infty}^{+\infty}\psi^*(x)\,\delta(x - x')\,dx = \boxed{\psi^*(x')}\]

Verification: Setting \(k=0\) in (b) gives \(\int \delta(x)dx = 1\), which is consistent with the normalization condition of the delta function. ✓

B-5. For the stationary state wave function, calculate the probability density and show that it is time-independent

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Solution Strategy: Directly calculate \(|\Psi(x,t)|^2 = \Psi^*(x,t)\Psi(x,t)\).

Detailed Calculation:

\[\Psi^*(x,t) = \psi^*(x)\,e^{+iEt/\hbar}\]

Therefore:

\[|\Psi(x,t)|^2 = \Psi^*\Psi = \psi^*(x)\,e^{+iEt/\hbar}\cdot\psi(x)\,e^{-iEt/\hbar}\]
\[= \psi^*(x)\psi(x)\cdot e^{+iEt/\hbar - iEt/\hbar} = |\psi(x)|^2 \cdot e^0 = |\psi(x)|^2\]

Final Answer:

\[\boxed{|\Psi(x,t)|^2 = |\psi(x)|^2}\]

The probability density does not depend on time \(t\). Since the absolute value of the phase factor \(e^{-iEt/\hbar}\) is 1, it does not contribute to the probability density. This is the reason these are called "stationary states."

Verification: It is essential that \(E\) is real. If \(E\) were complex, then \(|e^{-iEt/\hbar}|^2 \neq 1\), and probability would not be conserved. The Hermiticity of the Schrödinger equation guarantees that \(E\) is real. ✓

B-6. Hamiltonian Operator

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Solution Strategy: Calculate \(\hat{H}\psi\) in the region where \(V(x) = 0\) and confirm that it takes the form \(E\psi\).

Detailed Calculation:

Compute the second derivative of \(\psi(x) = Be^{-\kappa x}\):

\[\frac{d\psi}{dx} = -\kappa Be^{-\kappa x}\]
\[\frac{d^2\psi}{dx^2} = \kappa^2 Be^{-\kappa x} = \kappa^2\psi\]

Since \(V(x) = 0\):

\[\hat{H}\psi = -\frac{\hbar^2}{2m}\frac{d^2\psi}{dx^2} + 0 = -\frac{\hbar^2}{2m}\cdot\kappa^2\psi = -\frac{\hbar^2\kappa^2}{2m}\,\psi\]

This has the form \(\hat{H}\psi = E\psi\), so \(\psi\) is an eigenfunction of \(\hat{H}\).

Final Answer:

\[\boxed{E = -\frac{\hbar^2\kappa^2}{2m}}\]

Verification: \(E < 0\), which corresponds to a bound state (a wave function that decays exponentially outside the potential). Since \(\kappa > 0\), \(\psi = Be^{-\kappa x}\) converges to 0 as \(x \to +\infty\), which is physically reasonable. ✓ Also, dimensional analysis: \([\hbar^2\kappa^2/m] = [\text{J}\cdot\text{s}]^2[\text{m}^{-2}]/[\text{kg}] = [\text{J}]\) (dimensions of energy). ✓

B-7. Superposition of Two Energy Eigenstates

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Solution strategy: Expand \(|\Psi|^2 = \Psi^*\Psi\) and organize the cross terms.

Detailed calculation:

\[\Psi^* = \frac{1}{\sqrt{2}}\psi_1 e^{+iE_1 t/\hbar} + \frac{1}{\sqrt{2}}\psi_2 e^{+iE_2 t/\hbar}\]

(Since \(\psi_1, \psi_2\) are real, \(\psi_n^* = \psi_n\))

\[|\Psi|^2 = \Psi^*\Psi = \frac{1}{2}\left(\psi_1 e^{iE_1 t/\hbar} + \psi_2 e^{iE_2 t/\hbar}\right)\left(\psi_1 e^{-iE_1 t/\hbar} + \psi_2 e^{-iE_2 t/\hbar}\right)\]

Expanding:

\[= \frac{1}{2}\left[\psi_1^2 + \psi_2^2 + \psi_1\psi_2\,e^{i(E_1-E_2)t/\hbar} + \psi_1\psi_2\,e^{-i(E_1-E_2)t/\hbar}\right]\]
\[= \frac{1}{2}\left[\psi_1^2 + \psi_2^2 + 2\psi_1\psi_2\cos\!\left(\frac{(E_2-E_1)t}{\hbar}\right)\right]\]

Final answer:

\[\boxed{|\Psi(x,t)|^2 = \frac{1}{2}\psi_1^2 + \frac{1}{2}\psi_2^2 + \psi_1\psi_2\cos\!\left(\frac{(E_2-E_1)t}{\hbar}\right)}\]

The angular frequency of the oscillation in the interference term is:

\[\boxed{\omega_{12} = \frac{|E_2 - E_1|}{\hbar}}\]

Verification: When \(E_1 = E_2\), \(\omega_{12} = 0\) and there is no oscillation. This is consistent with the system being in a stationary state in the degenerate case. ✓ Furthermore, from Bohr's frequency condition \(\nu = (E_2 - E_1)/h\) and the relation \(\omega = 2\pi\nu\), we obtain \(\omega = (E_2 - E_1)/\hbar\), which agrees. ✓

B-8. For the momentum operator, write down explicitly. Furthermore, apply to the wave function and find the result.

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Solution strategy: Write out \(\hat{p}^2\) explicitly and compute the second derivative of the \(\sin\) function.

Detailed calculation:

\[\hat{p}^2 = \left(-i\hbar\frac{\partial}{\partial x}\right)^2 = (-i\hbar)^2\frac{\partial^2}{\partial x^2} = -\hbar^2\frac{\partial^2}{\partial x^2}\]

Acting on \(\psi(x) = A\sin(3\pi x/L)\):

\[\frac{\partial\psi}{\partial x} = A\cdot\frac{3\pi}{L}\cos\!\left(\frac{3\pi x}{L}\right)\]
\[\frac{\partial^2\psi}{\partial x^2} = -A\cdot\left(\frac{3\pi}{L}\right)^2\sin\!\left(\frac{3\pi x}{L}\right) = -\frac{9\pi^2}{L^2}\psi\]

Therefore:

\[\hat{p}^2\psi = -\hbar^2\cdot\left(-\frac{9\pi^2}{L^2}\right)\psi = \frac{9\pi^2\hbar^2}{L^2}\psi\]

Final answer:

\[\boxed{\hat{p}^2 = -\hbar^2\frac{\partial^2}{\partial x^2}}\]
\[\boxed{\hat{p}^2\psi = \frac{9\pi^2\hbar^2}{L^2}\,\psi}\]

\(\psi\) is an eigenfunction of \(\hat{p}^2\), with eigenvalue \(9\pi^2\hbar^2/L^2\).

Verification: This corresponds to the \(n=3\) state of an infinite square well (width \(L\)). The kinetic energy is \(\hat{p}^2/(2m) = 9\pi^2\hbar^2/(2mL^2)\), which matches the infinite well energy eigenvalue \(E_n = n^2\pi^2\hbar^2/(2mL^2)\) for \(n=3\). ✓ Note that \(\sin(3\pi x/L)\) is not an eigenfunction of \(\hat{p}\) (for the same reason as in D2(b)), but it is an eigenfunction of \(\hat{p}^2\). ✓

Medium

M-1. Derivation of the Time-Independent Schrödinger Equation via Separation of Variables

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Solution strategy: Substitute \(\Psi(x,t) = \psi(x)T(t)\) into the Schrödinger equation and use the separation of variables argument to obtain two ordinary differential equations.

Detailed calculation:

Substitute \(\Psi(x,t) = \psi(x)T(t)\) into equation (7.13):

\[i\hbar\frac{\partial}{\partial t}[\psi(x)T(t)] = -\frac{\hbar^2}{2m}\frac{\partial^2}{\partial x^2}[\psi(x)T(t)] + V(x)\psi(x)T(t)\]

Since \(\psi(x)\) does not depend on \(t\) and \(T(t)\) does not depend on \(x\):

\[i\hbar\,\psi(x)\frac{dT}{dt} = -\frac{\hbar^2}{2m}\frac{d^2\psi}{dx^2}\,T(t) + V(x)\psi(x)T(t)\]

Divide both sides by \(\psi(x)T(t)\):

\[i\hbar\frac{1}{T}\frac{dT}{dt} = -\frac{\hbar^2}{2m}\frac{1}{\psi}\frac{d^2\psi}{dx^2} + V(x)\]

The left-hand side is a function of \(t\) only, and the right-hand side is a function of \(x\) only. Since \(x\) and \(t\) are independent variables, for both sides to be equal, they must both equal the same constant. We denote this constant by \(E\).

The two ordinary differential equations obtained:

Time part:

\[i\hbar\frac{dT}{dt} = ET \tag{S1-1}\]

Spatial part:

\[-\frac{\hbar^2}{2m}\frac{d^2\psi}{dx^2} + V(x)\psi = E\psi \tag{S1-2}\]

Equation (S1-2) is the time-independent Schrödinger equation.

Solving the equation for \(T(t)\):

Rearranging equation (S1-1):

\[\frac{dT}{dt} = -\frac{iE}{\hbar}T\]

This is a first-order linear ordinary differential equation, and its solution is:

\[\boxed{T(t) = Ce^{-iEt/\hbar}}\]

Since the constant \(C\) can be absorbed into \(\psi(x)\), we can write \(T(t) = e^{-iEt/\hbar}\).

Therefore, the stationary-state wave function is:

\[\Psi(x,t) = \psi(x)\,e^{-iEt/\hbar}\]

Verification: - Physical meaning of the separation constant \(E\): Equation (S1-2) is \(\hat{H}\psi = E\psi\), so \(E\) is an energy eigenvalue. ✓ - \(T(t) = e^{-iEt/\hbar}\) satisfies \(|T|^2 = 1\), which is consistent with conservation of probability. ✓ - Dimensional analysis: \(Et/\hbar\) is dimensionless (\([E] = \text{J}\), \([t] = \text{s}\), \([\hbar] = \text{J}\cdot\text{s}\)). ✓

M-2. Derivation of Conservation of Probability (Continuity Equation)

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Solution strategy: Calculate \(\frac{\partial\rho}{\partial t}\), use the Schrödinger equation to replace \(\frac{\partial\Psi}{\partial t}\) and \(\frac{\partial\Psi^*}{\partial t}\), and show that the result equals \(-\frac{\partial j}{\partial x}\).

Detailed calculation:

Calculate the time derivative of the probability density \(\rho = \Psi^*\Psi\):

\[\frac{\partial\rho}{\partial t} = \frac{\partial\Psi^*}{\partial t}\Psi + \Psi^*\frac{\partial\Psi}{\partial t} \tag{S2-1}\]

From the Schrödinger equation (7.13):

\[\frac{\partial\Psi}{\partial t} = \frac{1}{i\hbar}\left[-\frac{\hbar^2}{2m}\frac{\partial^2\Psi}{\partial x^2} + V\Psi\right] \tag{S2-2}\]

Taking its complex conjugate (assuming \(V\) is real):

\[\frac{\partial\Psi^*}{\partial t} = -\frac{1}{i\hbar}\left[-\frac{\hbar^2}{2m}\frac{\partial^2\Psi^*}{\partial x^2} + V\Psi^*\right] = \frac{1}{-i\hbar}\left[-\frac{\hbar^2}{2m}\frac{\partial^2\Psi^*}{\partial x^2} + V\Psi^*\right] \tag{S2-3}\]

Substituting equations (S2-2) and (S2-3) into equation (S2-1):

\[\frac{\partial\rho}{\partial t} = \frac{1}{-i\hbar}\left[-\frac{\hbar^2}{2m}\frac{\partial^2\Psi^*}{\partial x^2} + V\Psi^*\right]\Psi + \Psi^*\frac{1}{i\hbar}\left[-\frac{\hbar^2}{2m}\frac{\partial^2\Psi}{\partial x^2} + V\Psi\right]\]

Organizing the \(V\) terms:

\[V\text{ contribution} = \frac{1}{-i\hbar}V\Psi^*\Psi + \frac{1}{i\hbar}V\Psi^*\Psi = -\frac{V|\Psi|^2}{i\hbar} + \frac{V|\Psi|^2}{i\hbar} = 0\]

The potential terms cancel. The remaining terms are:

\[\frac{\partial\rho}{\partial t} = \frac{1}{-i\hbar}\left(-\frac{\hbar^2}{2m}\right)\frac{\partial^2\Psi^*}{\partial x^2}\Psi + \frac{1}{i\hbar}\left(-\frac{\hbar^2}{2m}\right)\Psi^*\frac{\partial^2\Psi}{\partial x^2}\]
\[= \frac{\hbar}{2mi}\frac{\partial^2\Psi^*}{\partial x^2}\Psi - \frac{\hbar}{2mi}\Psi^*\frac{\partial^2\Psi}{\partial x^2}\]
\[= \frac{\hbar}{2mi}\left[\frac{\partial^2\Psi^*}{\partial x^2}\Psi - \Psi^*\frac{\partial^2\Psi}{\partial x^2}\right]\]

Here, we verify the following identity:

\[\frac{\partial}{\partial x}\left[\frac{\partial\Psi^*}{\partial x}\Psi - \Psi^*\frac{\partial\Psi}{\partial x}\right] = \frac{\partial^2\Psi^*}{\partial x^2}\Psi + \frac{\partial\Psi^*}{\partial x}\frac{\partial\Psi}{\partial x} - \frac{\partial\Psi^*}{\partial x}\frac{\partial\Psi}{\partial x} - \Psi^*\frac{\partial^2\Psi}{\partial x^2}\]
\[= \frac{\partial^2\Psi^*}{\partial x^2}\Psi - \Psi^*\frac{\partial^2\Psi}{\partial x^2}\]

Therefore:

\[\frac{\partial\rho}{\partial t} = \frac{\hbar}{2mi}\frac{\partial}{\partial x}\left[\frac{\partial\Psi^*}{\partial x}\Psi - \Psi^*\frac{\partial\Psi}{\partial x}\right]\]
\[= -\frac{\partial}{\partial x}\left[\frac{\hbar}{2mi}\left(\Psi^*\frac{\partial\Psi}{\partial x} - \frac{\partial\Psi^*}{\partial x}\Psi\right)\right]\]
\[= -\frac{\partial j}{\partial x}\]

Hence:

\[\boxed{\frac{\partial\rho}{\partial t} + \frac{\partial j}{\partial x} = 0}\]

Conservation of total probability:

\[\frac{d}{dt}\int_{-\infty}^{+\infty}\rho\,dx = \int_{-\infty}^{+\infty}\frac{\partial\rho}{\partial t}\,dx = -\int_{-\infty}^{+\infty}\frac{\partial j}{\partial x}\,dx = -\left[j(x)\right]_{-\infty}^{+\infty} = -(j(+\infty) - j(-\infty))\]

When \(\Psi\) goes to zero sufficiently rapidly as \(x \to \pm\infty\), we have \(j(\pm\infty) = 0\), so:

\[\boxed{\frac{d}{dt}\int_{-\infty}^{+\infty}|\Psi|^2\,dx = 0}\]

The total probability is independent of time.

Verification: The continuity equation has the same structure as charge conservation in electromagnetism, \(\frac{\partial\rho_e}{\partial t} + \nabla\cdot\mathbf{j}_e = 0\). It expresses the fact that probability can "flow" but cannot be "created or destroyed." ✓

M-3. Calculation of Probability Current Density

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Solution Strategy: Directly compute \(\rho\) and \(j\) for the plane wave \(\Psi = Ae^{i(kx-\omega t)}\).

(a) Probability density:

\[\rho = |\Psi|^2 = \Psi^*\Psi = Ae^{-i(kx-\omega t)}\cdot Ae^{i(kx-\omega t)} = A^2\]
\[\boxed{\rho = A^2}\]

(For real \(A\). Uniform and time-independent.)

(b) Probability current density:

Computing the necessary derivatives:

\[\frac{\partial\Psi}{\partial x} = ik\,\Psi, \quad \frac{\partial\Psi^*}{\partial x} = -ik\,\Psi^*\]
\[j = \frac{\hbar}{2mi}\left(\Psi^*\cdot ik\Psi - (-ik\Psi^*)\cdot\Psi\right)\]
\[= \frac{\hbar}{2mi}\left(ik|\Psi|^2 + ik|\Psi|^2\right)\]
\[= \frac{\hbar}{2mi}\cdot 2ik\,A^2 = \frac{\hbar}{2mi}\cdot 2ik\,A^2 = \frac{\hbar k}{m}A^2\]
\[\boxed{j = \frac{\hbar k}{m}A^2}\]

(c) Verification of \(j = \rho v\):

The particle velocity is \(v = p/m = \hbar k/m\). Therefore:

\[\rho v = A^2 \cdot \frac{\hbar k}{m} = \frac{\hbar k}{m}A^2 = j \quad \checkmark\]
\[\boxed{j = \rho v}\]

(d) Verification of the continuity equation:

\[\frac{\partial\rho}{\partial t} = \frac{\partial}{\partial t}(A^2) = 0\]
\[\frac{\partial j}{\partial x} = \frac{\partial}{\partial x}\left(\frac{\hbar k}{m}A^2\right) = 0\]

Therefore:

\[\frac{\partial\rho}{\partial t} + \frac{\partial j}{\partial x} = 0 + 0 = 0 \quad \checkmark\]

Sanity check: A plane wave has uniform probability density and uniform probability current. This represents a state where probability is "flowing uniformly," with no increase or decrease in probability anywhere, so the continuity equation is trivially satisfied. Physically, this is consistent with the picture of a particle with definite momentum flowing at constant velocity. ✓

M-4. Calculation of the Commutation Relation of Operators

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Solution Strategy: Apply \([\hat{x}, \hat{p}]\) to an arbitrary test function \(f(x)\) and show that the result is \(i\hbar f(x)\).

Detailed Calculation:

First, compute \(\hat{x}\hat{p}f(x)\):

\[\hat{x}\hat{p}f(x) = x\left(-i\hbar\frac{df}{dx}\right) = -i\hbar\,x\frac{df}{dx} \tag{S4-1}\]

Next, compute \(\hat{p}\hat{x}f(x)\):

\[\hat{p}\hat{x}f(x) = -i\hbar\frac{d}{dx}(xf) = -i\hbar\left(f + x\frac{df}{dx}\right) = -i\hbar f - i\hbar\,x\frac{df}{dx} \tag{S4-2}\]

(Using the product rule \(\frac{d}{dx}(xf) = f + x\frac{df}{dx}\))

Taking the difference:

\[[\hat{x}, \hat{p}]f(x) = \hat{x}\hat{p}f - \hat{p}\hat{x}f\]
\[= \left(-i\hbar\,x\frac{df}{dx}\right) - \left(-i\hbar f - i\hbar\,x\frac{df}{dx}\right)\]
\[= -i\hbar\,x\frac{df}{dx} + i\hbar f + i\hbar\,x\frac{df}{dx}\]
\[= i\hbar f(x)\]

Since \(f(x)\) is an arbitrary function, as an operator identity:

\[\boxed{[\hat{x}, \hat{p}] = i\hbar}\]

Verification: - Dimensional analysis: \([\hat{x}] = \text{m}\), \([\hat{p}] = \text{kg}\cdot\text{m/s}\), so \([\hat{x}\hat{p}] = \text{kg}\cdot\text{m}^2/\text{s} = [\hbar]\). ✓ - This commutation relation is the mathematical foundation of the uncertainty principle \(\Delta x\,\Delta p \geq \hbar/2\). - Verification with a specific example: For \(f(x) = e^{ikx}\), we have \(\hat{x}\hat{p}f = x\cdot\hbar k\,e^{ikx} = \hbar kx\,e^{ikx}\), and \(\hat{p}\hat{x}f = -i\hbar\frac{d}{dx}(xe^{ikx}) = -i\hbar(e^{ikx} + ikxe^{ikx}) = -i\hbar e^{ikx} + \hbar kx\,e^{ikx}\). The difference is \(\hbar kx\,e^{ikx} - (-i\hbar e^{ikx} + \hbar kx\,e^{ikx}) = i\hbar e^{ikx} = i\hbar f\). ✓

Advanced

A-1. Time Evolution of a Gaussian Wave Packet and Wave Packet Spreading

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(a) Fourier Transform

Solution strategy: Compute the Fourier transform of the Gaussian function using completing the square and the Gaussian integral formula.

Detailed calculation:

\[\phi(k) = \frac{1}{\sqrt{2\pi}}\int_{-\infty}^{+\infty}\Psi(x,0)\,e^{-ikx}\,dx\]
\[= \frac{1}{\sqrt{2\pi}}\left(\frac{1}{2\pi\sigma_0^2}\right)^{1/4}\int_{-\infty}^{+\infty}\exp\left(-\frac{x^2}{4\sigma_0^2} - ikx\right)dx\]

Complete the square in the exponent. Let \(a = \frac{1}{4\sigma_0^2}\), \(b = -ik\):

\[-ax^2 + bx = -a\left(x - \frac{b}{2a}\right)^2 + \frac{b^2}{4a}\]
\[= -\frac{1}{4\sigma_0^2}\left(x + 2i k\sigma_0^2\right)^2 + \frac{(-ik)^2}{4\cdot\frac{1}{4\sigma_0^2}}\]
\[= -\frac{1}{4\sigma_0^2}\left(x + 2ik\sigma_0^2\right)^2 - k^2\sigma_0^2\]

Applying the Gaussian integral formula \(\int_{-\infty}^{+\infty}e^{-a(x-x_0)^2}dx = \sqrt{\pi/a}\) (valid for complex shifts as long as \(\text{Re}(a) > 0\)):

\[\int_{-\infty}^{+\infty}\exp\left(-\frac{x^2}{4\sigma_0^2} - ikx\right)dx = e^{-k^2\sigma_0^2}\cdot\sqrt{\frac{\pi}{1/(4\sigma_0^2)}} = e^{-k^2\sigma_0^2}\cdot 2\sigma_0\sqrt{\pi}\]

Therefore:

\[\phi(k) = \frac{1}{\sqrt{2\pi}}\left(\frac{1}{2\pi\sigma_0^2}\right)^{1/4}\cdot 2\sigma_0\sqrt{\pi}\,e^{-k^2\sigma_0^2}\]

Simplifying:

\[\phi(k) = \frac{2\sigma_0\sqrt{\pi}}{\sqrt{2\pi}}\left(\frac{1}{2\pi\sigma_0^2}\right)^{1/4}e^{-k^2\sigma_0^2}\]
\[= \sqrt{2\sigma_0^2}\cdot\frac{\sqrt{\pi}}{\sqrt{\pi}}\left(\frac{1}{2\pi\sigma_0^2}\right)^{1/4}e^{-k^2\sigma_0^2}\]

Computing more carefully:

\[\frac{2\sigma_0\sqrt{\pi}}{\sqrt{2\pi}} = \frac{2\sigma_0}{\sqrt{2}} = \sigma_0\sqrt{2}\]
\[\phi(k) = \sigma_0\sqrt{2}\cdot\left(\frac{1}{2\pi\sigma_0^2}\right)^{1/4}e^{-k^2\sigma_0^2}\]
\[= \sigma_0\sqrt{2}\cdot\frac{1}{(2\pi)^{1/4}\sigma_0^{1/2}}e^{-k^2\sigma_0^2}\]
\[= \frac{\sqrt{2}\,\sigma_0^{1/2}}{(2\pi)^{1/4}}e^{-k^2\sigma_0^2}\]

Rewriting this in the standard Gaussian form. To verify the form of \(\phi(k)\):

\[\phi(k) = \left(\frac{2\sigma_0^2}{\pi}\right)^{1/4}e^{-k^2\sigma_0^2}\]

Verification: \(\left(\frac{2\sigma_0^2}{\pi}\right)^{1/4} = \frac{(2\sigma_0^2)^{1/4}}{\pi^{1/4}} = \frac{2^{1/4}\sigma_0^{1/2}}{\pi^{1/4}}\)

On the other hand, the previous result gives \(\frac{\sqrt{2}\,\sigma_0^{1/2}}{(2\pi)^{1/4}} = \frac{2^{1/2}\sigma_0^{1/2}}{2^{1/4}\pi^{1/4}} = \frac{2^{1/4}\sigma_0^{1/2}}{\pi^{1/4}}\). These agree. ✓

Final answer:

\[\boxed{\phi(k) = \left(\frac{2\sigma_0^2}{\pi}\right)^{1/4}\exp(-k^2\sigma_0^2)}\]

This is a Gaussian function in \(k\)-space, with a width in momentum space of \(\Delta k = 1/(2\sigma_0)\) (i.e., \(\Delta p = \hbar/(2\sigma_0)\)).

(b) Wave function at time \(t\)

Solution strategy: Perform the inverse Fourier transform using \(\omega = \hbar k^2/(2m)\).

Detailed calculation:

\[\Psi(x,t) = \frac{1}{\sqrt{2\pi}}\int_{-\infty}^{+\infty}\phi(k)\,e^{i(kx - \omega t)}\,dk\]
\[= \frac{1}{\sqrt{2\pi}}\left(\frac{2\sigma_0^2}{\pi}\right)^{1/4}\int_{-\infty}^{+\infty}\exp\left(-k^2\sigma_0^2 + ikx - i\frac{\hbar k^2}{2m}t\right)dk\]

Collecting terms in the exponent:

\[-k^2\sigma_0^2 - i\frac{\hbar t}{2m}k^2 + ikx = -k^2\left(\sigma_0^2 + \frac{i\hbar t}{2m}\right) + ikx\]

Define \(\alpha(t) \equiv \sigma_0^2 + \frac{i\hbar t}{2m}\):

\[\text{Exponent} = -\alpha k^2 + ikx\]

Completing the square:

\[-\alpha k^2 + ikx = -\alpha\left(k - \frac{ix}{2\alpha}\right)^2 - \frac{x^2}{4\alpha}\]

Performing the Gaussian integral (\(\text{Re}(\alpha) = \sigma_0^2 > 0\)):

\[\int_{-\infty}^{+\infty}\exp\left[-\alpha\left(k - \frac{ix}{2\alpha}\right)^2\right]dk = \sqrt{\frac{\pi}{\alpha}}\]

Therefore:

\[\Psi(x,t) = \frac{1}{\sqrt{2\pi}}\left(\frac{2\sigma_0^2}{\pi}\right)^{1/4}\sqrt{\frac{\pi}{\alpha}}\,\exp\left(-\frac{x^2}{4\alpha}\right)\]
\[= \left(\frac{2\sigma_0^2}{\pi}\right)^{1/4}\frac{1}{\sqrt{2\alpha}}\,\exp\left(-\frac{x^2}{4\alpha}\right)\]

Computing the probability density:

\[|\Psi(x,t)|^2 = \left(\frac{2\sigma_0^2}{\pi}\right)^{1/2}\frac{1}{2|\alpha|}\,\exp\left(-\frac{x^2}{4}\cdot 2\text{Re}\!\left(\frac{1}{\alpha}\right)\right)\]

Computing \(\frac{1}{\alpha}\):

\[\frac{1}{\alpha} = \frac{1}{\sigma_0^2 + i\hbar t/(2m)} = \frac{\sigma_0^2 - i\hbar t/(2m)}{\sigma_0^4 + (\hbar t/(2m))^2}\]
\[\text{Re}\!\left(\frac{1}{\alpha}\right) = \frac{\sigma_0^2}{\sigma_0^4 + (\hbar t/(2m))^2} = \frac{1}{\sigma_0^2\left(1 + (\hbar t/(2m\sigma_0^2))^2\right)} = \frac{1}{\sigma_0^2 + (\hbar t)^2/(4m^2\sigma_0^2)}\]

Define \(\sigma(t)^2 \equiv \sigma_0^2 + \frac{\hbar^2 t^2}{4m^2\sigma_0^2} = \sigma_0^2\left(1 + \left(\frac{\hbar t}{2m\sigma_0^2}\right)^2\right)\):

\[\text{Re}\!\left(\frac{1}{\alpha}\right) = \frac{1}{\sigma(t)^2}\]

Also, \(|\alpha|^2 = \sigma_0^4 + (\hbar t/(2m))^2 = \sigma_0^2\cdot\sigma(t)^2\), so \(|\alpha| = \sigma_0\,\sigma(t)\).

Therefore:

\[|\Psi(x,t)|^2 = \left(\frac{2\sigma_0^2}{\pi}\right)^{1/2}\frac{1}{2\sigma_0\sigma(t)}\,\exp\left(-\frac{x^2}{2\sigma(t)^2}\right)\]
\[= \frac{1}{\sqrt{2\pi}\,\sigma(t)}\,\exp\left(-\frac{x^2}{2\sigma(t)^2}\right)\]

(Verification: \(\left(\frac{2\sigma_0^2}{\pi}\right)^{1/2}\frac{1}{2\sigma_0\sigma(t)} = \frac{\sqrt{2}\sigma_0}{\sqrt{\pi}}\cdot\frac{1}{2\sigma_0\sigma(t)} = \frac{1}{\sqrt{2\pi}\sigma(t)}\). ✓)

This is a Gaussian distribution with mean 0 and standard deviation \(\sigma(t)\), confirming that the probability density remains Gaussian.

(c) Spreading of the wave packet

Final answer:

\[\boxed{\sigma(t) = \sigma_0\sqrt{1 + \left(\frac{\hbar t}{2m\sigma_0^2}\right)^2}}\]

Discussion of physical meaning:

When the initial position uncertainty is \(\Delta x = \sigma_0\), the uncertainty principle \(\Delta x\,\Delta p \geq \hbar/2\) requires a momentum uncertainty of at least:

\[\Delta p \geq \frac{\hbar}{2\sigma_0}\]

(The Gaussian wave packet is a minimum uncertainty state, so equality holds.)

This momentum uncertainty implies a velocity uncertainty \(\Delta v = \Delta p/m = \hbar/(2m\sigma_0)\). As time \(t\) elapses, the spatial spread increases due to the velocity uncertainty:

\[\Delta x(t) \sim \sqrt{\sigma_0^2 + (\Delta v\cdot t)^2} = \sigma_0\sqrt{1 + \left(\frac{\hbar t}{2m\sigma_0^2}\right)^2}\]

This is precisely the expression for \(\sigma(t)\).

  • Short times (\(t \ll 2m\sigma_0^2/\hbar\)): \(\sigma(t) \approx \sigma_0\) (almost no spreading)
  • Long times (\(t \gg 2m\sigma_0^2/\hbar\)): \(\sigma(t) \approx \frac{\hbar t}{2m\sigma_0}\) (linear spreading)
  • Making \(\sigma_0\) smaller gives a more precisely determined initial position, but \(\Delta p\) becomes larger, causing the wave packet to spread more rapidly. This is a direct manifestation of the trade-off inherent in the uncertainty principle.

Verification: - At \(t = 0\): \(\sigma(0) = \sigma_0\). ✓ - In the classical limit \(\hbar \to 0\): \(\sigma(t) = \sigma_0\) (no spreading). This is consistent with classical particles having definite trajectories. ✓ - Normalization of \(|\Psi(x,t)|^2\): \(\int_{-\infty}^{+\infty}\frac{1}{\sqrt{2\pi}\sigma(t)}e^{-x^2/(2\sigma(t)^2)}dx = 1\). ✓ (normalization condition for a Gaussian distribution)

A-2. Ehrenfest's Theorem

→ Back to problem

(i) Derivation of \(\frac{d\langle x\rangle}{dt} = \frac{\langle p\rangle}{m}\)

Strategy: Compute the time derivative of \(\langle x\rangle\), substitute \(\partial\Psi/\partial t\) using the Schrödinger equation, and perform integration by parts.

Detailed calculation:

\[\frac{d\langle x\rangle}{dt} = \frac{d}{dt}\int_{-\infty}^{+\infty}\Psi^* x\,\Psi\,dx = \int_{-\infty}^{+\infty}\left(\frac{\partial\Psi^*}{\partial t}x\Psi + \Psi^* x\frac{\partial\Psi}{\partial t}\right)dx\]

From the Schrödinger equation:

\[\frac{\partial\Psi}{\partial t} = \frac{1}{i\hbar}\hat{H}\Psi = \frac{1}{i\hbar}\left(-\frac{\hbar^2}{2m}\frac{\partial^2\Psi}{\partial x^2} + V\Psi\right)\]
\[\frac{\partial\Psi^*}{\partial t} = -\frac{1}{i\hbar}\hat{H}^*\Psi^* = -\frac{1}{i\hbar}\left(-\frac{\hbar^2}{2m}\frac{\partial^2\Psi^*}{\partial x^2} + V\Psi^*\right)\]

Substituting:

\[\frac{d\langle x\rangle}{dt} = \int\left[-\frac{1}{i\hbar}\left(-\frac{\hbar^2}{2m}\Psi^*_{xx} + V\Psi^*\right)x\Psi + \Psi^* x\frac{1}{i\hbar}\left(-\frac{\hbar^2}{2m}\Psi_{xx} + V\Psi\right)\right]dx\]

The \(V\) terms:

\[\frac{1}{i\hbar}\left[-V\Psi^* x\Psi + \Psi^* x V\Psi\right] = 0\]

(These cancel because \(V\) is real and commutes with \(x\).)

The remaining kinetic energy terms:

\[\frac{d\langle x\rangle}{dt} = \frac{\hbar}{2mi}\int\left[\Psi^*_{xx}\cdot x\Psi - \Psi^*\cdot x\Psi_{xx}\right]dx\]

We perform integration by parts. Integrate \(\int \Psi^*_{xx}\cdot x\Psi\,dx\) by parts twice.

First, integrate \(\int \frac{\partial^2\Psi^*}{\partial x^2}\,x\Psi\,dx\) by parts (transferring derivatives from \(\Psi^*_{xx}\)):

\[\int \frac{\partial^2\Psi^*}{\partial x^2}\,x\Psi\,dx = \left[\frac{\partial\Psi^*}{\partial x}\,x\Psi\right]_{-\infty}^{+\infty} - \int\frac{\partial\Psi^*}{\partial x}\frac{\partial}{\partial x}(x\Psi)\,dx\]

The boundary term vanishes since \(\Psi\) goes to 0 at infinity.

\[= -\int\frac{\partial\Psi^*}{\partial x}\left(\Psi + x\frac{\partial\Psi}{\partial x}\right)dx = -\int\Psi^*_x\Psi\,dx - \int\Psi^*_x\,x\Psi_x\,dx\]

Similarly, integrate \(\int\Psi^*\,x\Psi_{xx}\,dx\) by parts:

\[\int\Psi^*\,x\Psi_{xx}\,dx = \left[\Psi^*\,x\Psi_x\right]_{-\infty}^{+\infty} - \int\frac{\partial}{\partial x}(\Psi^* x)\,\Psi_x\,dx\]
\[= -\int(\Psi^*_x\,x + \Psi^*)\Psi_x\,dx = -\int\Psi^*_x\,x\Psi_x\,dx - \int\Psi^*\Psi_x\,dx\]

Taking the difference:

\[\int\Psi^*_{xx}\,x\Psi\,dx - \int\Psi^*\,x\Psi_{xx}\,dx\]
\[= \left(-\int\Psi^*_x\Psi\,dx - \int\Psi^*_x\,x\Psi_x\,dx\right) - \left(-\int\Psi^*_x\,x\Psi_x\,dx - \int\Psi^*\Psi_x\,dx\right)\]
\[= -\int\Psi^*_x\Psi\,dx + \int\Psi^*\Psi_x\,dx\]

Therefore:

\[\frac{d\langle x\rangle}{dt} = \frac{\hbar}{2mi}\left(-\int\Psi^*_x\Psi\,dx + \int\Psi^*\Psi_x\,dx\right)\]
\[= \frac{\hbar}{2mi}\cdot 2\int\Psi^*\frac{\partial\Psi}{\partial x}\,dx\]

(Here we used \(\int\Psi^*_x\Psi\,dx = -\int\Psi^*\Psi_x\,dx\), which follows from integration by parts: \(\int\Psi^*_x\Psi\,dx = [\Psi^*\Psi]_{-\infty}^{+\infty} - \int\Psi^*\Psi_x\,dx = -\int\Psi^*\Psi_x\,dx\).)

Thus:

\[\frac{d\langle x\rangle}{dt} = \frac{\hbar}{mi}\int\Psi^*\frac{\partial\Psi}{\partial x}\,dx\]

Here \(\frac{\hbar}{mi}\cdot\frac{\partial}{\partial x} = \frac{1}{m}\cdot\frac{\hbar}{i}\frac{\partial}{\partial x} = \frac{1}{m}\cdot(-1)\cdot(-i\hbar)\frac{\partial}{\partial x}\)...

More directly:

\[\frac{d\langle x\rangle}{dt} = \frac{1}{m}\int\Psi^*\left(-i\hbar\frac{\partial}{\partial x}\right)\Psi\,dx = \frac{1}{m}\int\Psi^*\hat{p}\Psi\,dx = \frac{\langle p\rangle}{m}\]

(Verification that \(\frac{\hbar}{mi} = \frac{1}{m}\cdot\frac{\hbar}{i} = \frac{1}{m}\cdot(-i\hbar)\): \(\frac{\hbar}{mi}\frac{\partial\Psi}{\partial x} = \frac{1}{m}\cdot\frac{\hbar}{i}\frac{\partial\Psi}{\partial x} = \frac{1}{m}\cdot(-i\hbar)\frac{\partial\Psi}{\partial x} = \frac{\hat{p}\Psi}{m}\). ✓)

\[\boxed{\frac{d\langle x\rangle}{dt} = \frac{\langle p\rangle}{m}}\]

(ii) Derivation of \(\frac{d\langle p\rangle}{dt} = -\left\langle\frac{dV}{dx}\right\rangle\)

Detailed calculation:

\[\frac{d\langle p\rangle}{dt} = \frac{d}{dt}\int\Psi^*\left(-i\hbar\frac{\partial}{\partial x}\right)\Psi\,dx = -i\hbar\int\left(\frac{\partial\Psi^*}{\partial t}\frac{\partial\Psi}{\partial x} + \Psi^*\frac{\partial}{\partial x}\frac{\partial\Psi}{\partial t}\right)dx\]

Substituting the Schrödinger equation:

\[\frac{\partial\Psi}{\partial t} = \frac{1}{i\hbar}\left(-\frac{\hbar^2}{2m}\Psi_{xx} + V\Psi\right)\]

Computing the second term:

\[-i\hbar\int\Psi^*\frac{\partial}{\partial x}\left[\frac{1}{i\hbar}\left(-\frac{\hbar^2}{2m}\Psi_{xx} + V\Psi\right)\right]dx\]
\[= -i\hbar\cdot\frac{1}{i\hbar}\int\Psi^*\frac{\partial}{\partial x}\left(-\frac{\hbar^2}{2m}\Psi_{xx} + V\Psi\right)dx\]
\[= -\int\Psi^*\left(-\frac{\hbar^2}{2m}\Psi_{xxx} + V_x\Psi + V\Psi_x\right)dx\]

Computing the first term:

\[-i\hbar\int\frac{\partial\Psi^*}{\partial t}\Psi_x\,dx = -i\hbar\int\left[-\frac{1}{i\hbar}\left(-\frac{\hbar^2}{2m}\Psi^*_{xx} + V\Psi^*\right)\right]\Psi_x\,dx\]
\[= \int\left(-\frac{\hbar^2}{2m}\Psi^*_{xx} + V\Psi^*\right)\Psi_x\,dx\]

Combining:

\[\frac{d\langle p\rangle}{dt} = \int\left(-\frac{\hbar^2}{2m}\Psi^*_{xx}\Psi_x + V\Psi^*\Psi_x\right)dx + \int\Psi^*\left(\frac{\hbar^2}{2m}\Psi_{xxx} - V_x\Psi - V\Psi_x\right)dx\]

The \(V\Psi^*\Psi_x\) term and the \(-\Psi^* V\Psi_x\) term cancel:

\[\frac{d\langle p\rangle}{dt} = \frac{\hbar^2}{2m}\int\left(\Psi^*\Psi_{xxx} - \Psi^*_{xx}\Psi_x\right)dx - \int\Psi^*\frac{dV}{dx}\Psi\,dx\]

We show that the first term vanishes. Using integration by parts:

\[\int\Psi^*\Psi_{xxx}\,dx = [\Psi^*\Psi_{xx}]_{-\infty}^{+\infty} - \int\Psi^*_x\Psi_{xx}\,dx = -\int\Psi^*_x\Psi_{xx}\,dx\]
\[\int\Psi^*_{xx}\Psi_x\,dx = [\Psi^*_x\Psi_x]_{-\infty}^{+\infty} - \int\Psi^*_x\Psi_{xx}\,dx = -\int\Psi^*_x\Psi_{xx}\,dx\]

(All boundary terms vanish.)

Therefore \(\int\Psi^*\Psi_{xxx}\,dx - \int\Psi^*_{xx}\Psi_x\,dx = 0\).

Thus:

\[\boxed{\frac{d\langle p\rangle}{dt} = -\left\langle\frac{dV}{dx}\right\rangle}\]

Correspondence with Newton's Equation of Motion

Combining the two results:

\[\frac{d\langle x\rangle}{dt} = \frac{\langle p\rangle}{m}, \quad \frac{d\langle p\rangle}{dt} = -\left\langle\frac{dV}{dx}\right\rangle\]

The first equation states "the rate of change of the expectation value of position = expectation value of momentum / mass," corresponding to the classical relation \(v = p/m\).

The second equation states "the rate of change of the expectation value of momentum = expectation value of the force," which is the quantum mechanical counterpart of Newton's second law \(F = dp/dt = -dV/dx\).

Combined, these give:

\[m\frac{d^2\langle x\rangle}{dt^2} = -\left\langle\frac{dV}{dx}\right\rangle\]

This demonstrates that "expectation values obey Newton's equation of motion."

The Case Where \(V(x)\) is a Polynomial of Degree 2 or Less

When \(V(x)\) is a polynomial of degree 2 or less in \(x\) (constant, linear potential \(V = mgx\), harmonic oscillator \(V = \frac{1}{2}m\omega^2 x^2\)):

\[\frac{dV}{dx} = ax + b \quad (\text{a function linear or lower in } x)\]

In this case:

\[\left\langle\frac{dV}{dx}\right\rangle = a\langle x\rangle + b = \frac{dV}{dx}\bigg|_{x = \langle x\rangle}\]

That is, the expectation value of the force equals "the force evaluated at the expectation value of position." This holds exactly because \(dV/dx\) is at most linear in \(x\), so \(\langle f(x)\rangle = f(\langle x\rangle)\) is strictly satisfied (the case where Jensen's inequality becomes an equality).

In this case, the equation of motion for expectation values becomes:

\[m\frac{d^2\langle x\rangle}{dt^2} = -\frac{dV}{dx}\bigg|_{x = \langle x\rangle}\]

This has exactly the same form as the classical equation of motion, and the expectation value \(\langle x\rangle\) follows the same trajectory as a classical particle \(x_{\text{cl}}(t)\).

For a general potential, \(\left\langle\frac{dV}{dx}\right\rangle \neq \frac{dV}{dx}\big|_{\langle x\rangle}\) (due to contributions from higher-order terms), and quantum corrections appear. When the wave packet spread is small (\(\Delta x\) is small compared to the characteristic scale of the potential), the motion approximately follows the classical trajectory—this is one aspect of the transition to the classical limit.

Verification: - Free particle (\(V = 0\)): \(\frac{d\langle p\rangle}{dt} = 0\) (momentum conservation), \(\langle x\rangle = \langle x\rangle_0 + \frac{\langle p\rangle}{m}t\) (uniform rectilinear motion). Consistent with classical mechanics. ✓ - Harmonic oscillator (\(V = \frac{1}{2}m\omega^2 x^2\)): \(\frac{d\langle p\rangle}{dt} = -m\omega^2\langle x\rangle\). \(\langle x\rangle\) oscillates with angular frequency \(\omega\). Consistent with classical harmonic oscillation. ✓ - Dimensional analysis: \([d\langle p\rangle/dt] = [\text{force}]\), \([\langle dV/dx\rangle] = [\text{energy}/\text{length}] = [\text{force}]\). ✓